你需要 javascript ajax和json以及回调函数来不断检查.它可以通过推送服务完成,但 PHP很糟糕,你需要websockest等.facebook使用超时调用.不确定stackoverflow

refreshInterval = 500
  refreshTimeout = setTimeout( getNotificationCounts,refreshInterval );
function getNotifications() {
        $.ajax({
            url : 'path_to_your_PHP',tyPE : 'POST',data:   //put for exampel user_id but IT cvan be handled by sesion as well
            dataType : 'json',success : function(data) {
                alert('You have'+ data.total +' new messages')
                refreshTimeout = setTimeout( getNotificationCounts,refreshInterval ); //this will check every half of second
            }
        });
    }

然后在PHP中,例如,您在数据库中有一个标志,您可以使用该标志检查数据是否是新数据库.注意格式化和正确的sql转义

$result=MysqL_query("SELECT count(*) as total From myTable where user_id=$_POST['user_id'] AND is_read=0"); 

      //note instad of $_POST you can use session if users are LOGged in,or however you are handling it
        $data=MysqL_fetch_assoc($result);
        echo json_encode($data)

;

之后,您可以创建另一个ajax调用,在用户单击它们时将其标记为已读.或打开对话框列表或wahtever

MysqL_query("UPDATE myTable WHERE user_id=$_POST['user_id'] SET  is_read=1");